Thermodynamics Test

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Thermodynamics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Specific heat of iron at 25°C and 1.00 bar is

A
0.4454 JK^-1 g^-1

B
24.942 JK^-1 mol^-1

C
0.1071 cal K^-1 g^-1

D
All are correct
Question 2
1 / -0

A rather soft, silvery metal was observed to have a specific heat of 0.225 cal K^-1g^-1. Thus, metal is

A
Na

B
Mg

C
Al

D
Fe

Solution

It should be better to remove this question. It's data based.
Question 3
1 / -0

Specific heat of water is 4.184 JK^-1g^-1, Rise in temperature when 1 kJ of heat is absorbed by 2 moles of H₂O is

A
119.5^o

B
0.37^o

C
0.12^o

D
6.64^o

Solution

We have,
Q = ms∆T
1000 = 2*18*4.184*∆T
∆T = 6.63 K
Question 4
1 / -0

The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is

A
– 100 x kJ/mol

B
– 50 x kJ/mol

C
+ 100 x kJ/mol

D
+50 x kJ/mol

Solution
Question 5
1 / -0

The heat capacity ratio, r was determined for cyanogen as 1.177. Thus, C_p for this gas is

A
55.3 JK^-1 mol^-1

B
33.26 JK^-1 mol^-1

C
8.314 JK^-1 mol^-1

D
24.94 JK^-1 mol^-1

Solution

γ = 1.177
C_V = R/γ-1
= 8.314/0.177 = 46.97
Also γ = C_P/C_V
1.177 = C_P/46.97
Or C_P= 55.28
Question 6
1 / -0

Assuming the composition of air to be X (N₂) = 0.80, X (O₂) = 0.18 and X (CO₂) = 0.02, molar heat capacity of air at constant pressure is

A
3.50R

B
4.00R

C
2.50R

D
3.51R

Solution

(C_p)_mixture = (µ₁C_p1+µ₂C_p2+µ₃C_p3+.........)/µ₁+µ₂+µ₃+......... where µ is the no of moles.
Or (C_p)_mixture = ƞ₁C_p1+ƞ₂C_p2+ƞ₃C_p3+............. Where ƞ is the corr. mole fraction.
On putting the values, we have
Cp_mixture= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
Question 7
1 / -0

Latent heat of fusion of ice is 6.02 kJ mol^-1. The heat capacity of water is 4.18 Jg^-1K^-1. 500 g of liquid water is to be cooled from 20°C to 0°C . Number of ice cubes (each of one mole) required is

A
1

B
7

C
14

D
125

Solution

Molecular mass of hydrogen = 1
Molecular mass of oxygen =16
Molecular mass of water = 18 gram per mole
As,
n = m / M
n = 500 / 18
n = 27.8 moles
For cooling 27.8 moles at 20 = 27.8 x 75.4 x 20 = 41.922 kilo joule
41.922/6 = 6.987 that is approximately equal to 7
So, the number of ice cubes that are required to cool the water will be 7.
Question 8
1 / -0

Sulphur (2.56 g) is burned in a constant volume calorimeter with excess O₂(g). The temperature increases from 21.25°C to 26.72°C . The bomb has a heat capacity of 923 JK^-1. Calorimeter contains 815 g of water. Thus, change in internal energy per mole of SO₂ formed for the reaction is
(specific heat of water is 4.184 JK^-1g^-1.)

A
- 296.27 kJ

B
+ 296.27 kJ

C
- 2370.13 kJ

D
+ 2370.13 kJ

Solution

Moles of S8 = 2.56/256 = 0.01
So, moles of SO₂ formed = 0.08
Rise in temperature = (273+26.72)K - (273+21.25)K
= 5.47 K
Total Energy in internal energy (as system is at constant volume) = -(923 jK-1 × 5.47 +815g×4.18)
= -23701.29 J = -23.70129 kJ
Thus change in internal energy per mole of formed = -23.7012kJ/0.08 = -296.3 kJ
Question 9
1 / -0

Direction (Q. Nos. 9) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : C_v value of helium (He) is always 3/2R but C_v value of hydrogen (H₂) is 3/2R at low temperature, 5/2R at moderate temperature and more than 5/2R at higher temperature.
Statement II : At lower temperature, only translational degree of freedom contributes to heat capacity while at higher temperature, rotational and vibrational degrees of freedom also contribute to heat capacity.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

In case of helium (monoatomic gas) we have only three degrees of freedom which correspond to three translational motion so the total heat capacity will increase. The contributors by vibrational motion are not appreciable at low temperature but increase from 0 to R when temperature increases.
CV=−f(R)/2
CV=-f(R)2, where f is the degree of freedom. At low temperature only translational motion is considered and f=3
∴CV=3R/2
f=3.∴CV=3R2 At moderate temperature both translational and rotational motions are considered.
f=3+2
f=3+2 (3-translational and 2 rotational).
∴CV=5R/2
∴CV=5R2. At still high temperature translational, rotational and vibrational motions are considered.
f=3+2+2
f=3+2+2 (3-translatinal , 2-rotational, 2-vibrational).
∴CV=7R/2
Question 10
1 / -0

Direction (Q. Nos. 10) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q. According to the classical equipartition of energy theory, the contributions to the molar heat capacity at constant volume are for each degree of translational freedom, for each degree of rotational freedom and (R) for each degree of translational freedom. Match the species in Column I with the values of molar heat capacity (C_p or C_V) in Column II and select answer from codes given below the table.

A
a

B
b

C
c

D
d

Solution

For monoatomic gas, C_P = 5R/2
For diatomic gas, C_P = 9R/2
For triatomic gas, C_V = 6R and C_P = C_V+R = 7R

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Thermodynamics Test - 18

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Thermodynamics Test - 18

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Question 1

0.4454 JK-1 g-1

24.942 JK-1 mol-1

0.1071 cal K-1 g-1

All are correct

Question 2

Na

Mg

Al

Fe

Solution

Question 3

119.5o

0.37o

0.12o

6.64o

Solution

Question 4

– 100 x kJ/mol

– 50 x kJ/mol

+ 100 x kJ/mol

+50 x kJ/mol

Solution

Question 5

55.3 JK-1 mol-1

33.26 JK-1 mol-1

8.314 JK-1 mol-1

24.94 JK-1 mol-1

Solution

Question 6

3.50R

4.00R

2.50R

3.51R

Solution

Question 7

1

7

14

125

Solution

Question 8

- 296.27 kJ

+ 296.27 kJ

- 2370.13 kJ

+ 2370.13 kJ

Solution

Question 9

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

Statement I is correct but Statement II is incorrect

Statement II is correct but Statement I is incorrect

Solution

Question 10

a

b

c

d

Solution

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0.4454 JK^-1 g^-1

24.942 JK^-1 mol^-1

0.1071 cal K^-1 g^-1

119.5^o

0.37^o

0.12^o

6.64^o

55.3 JK^-1 mol^-1

33.26 JK^-1 mol^-1

8.314 JK^-1 mol^-1

24.94 JK^-1 mol^-1